A cylindrical rod having temperature `T_(1)` and `T_(2)` at its ends. The rate of flow of heat is `Q_(1) cal//sec`. If all the linear dimensions are doubled keeping temperature constant, then rate of flow of heat `Q_(2)` will be

Options

A. `4 Q_(1)`
B. `2 Q_(1)`
C. `(Q_(1))/(4)`
D. `(Q_(1))/(2)`

Answer

B

Explanation

(2) `(Q_(1))/(Q_(2)) = (R_(2))/(R_(1)) = (2 l //K . Pi (2 r)^(2))/(l..K . Pi r^(2)) = 1//2`
`Q_(2) = 2Q_(1)`